An interesting theorem: If \(z + \dfrac1z = 2\cos \alpha\), then \(z^n + \dfrac1{z^n} = 2\cos n\alpha\) for positive integers n and real \(\alpha\).
(If you don't like to read proofs, then you can jump to the point where I marked "END OF PROOF".)
Proof of the theorem:
We use strong induction.
Base case #1 : n = 1.
Substituting n = 1 and checking shows that base case #1 is true.
Base case #2 : n = 2.
\(z^2 + \dfrac1{z^2} = \left(z + \dfrac1z\right)^2 - 2 = (2\cos \alpha)^2 - 2=2(2\cos^2 \alpha - 1) = 2\cos 2\alpha\)
So base case #2 is true.
Inductive step: Assume \(z^t + \dfrac1{z^t} = 2\cos t\alpha\) and \(z^{t - 1} + \dfrac1{z^{t - 1}} = 2\cos(t -1)\alpha\) for some positive integer \(t\).
\(\quad z^{t + 1} + \dfrac1{z^{t + 1}}\\ = \left(z^t + \dfrac1{z^t}\right)\left(z + \dfrac1z\right) - \left(z^{t - 1} + \dfrac1{z^{t - 1}}\right)\\ = 2\cos t\alpha \cdot 2\cos \alpha - 2\cos (t - 1)\alpha\\ \text{By product to sum formula:}\\ = 2 \left(\cos (t + 1)\alpha + \cos(t - 1)\alpha\right) - 2\cos(t-1)\alpha\\ =2\cos(t + 1)\alpha\)
Therefore by principles of mathematical induction, the proposition is true.
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Back to the problem. Using the theorem I mentioned above, and substituting \(\alpha = 60^\circ\) into the formula,
\(\quad z^{10} + \dfrac1{z^{10}}\\ = 2\cos\left(10\cdot 60^\circ\right)\\ = 2 \cos\left(600^\circ\right)\\ = 2 \cos 240^\circ\\ = \boxed{-1}\)
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